Question

1. if (lm = 22) and (mn = 15), find (ln).
2. if (ln = 54) and (lm = 31), find (mn).
3. if (df=9x - 39), find (ef).

if (hj = 7x - 27), find the value of (x).
if (cd = 5x - 9), find each value.
1, and (ac = 43), find (ab).

Explanation:

Step1: Use segment - addition postulate

For the first problem, if points $L$, $M$, and $N$ are collinear, then $LN=LM + MN$. Given $LM = 22$ and $MN=15$, we substitute the values into the formula: $LN=22 + 15$.
$LN=37$

Step2: Use segment - addition postulate for the second problem

If $LN=LM + MN$, then $MN=LN - LM$. Given $LN = 54$ and $LM = 31$, we substitute the values: $MN=54 - 31$.
$MN = 23$

Step3: Solve for $x$ in the third problem

Since $DE+EF=DF$ and $DE = 47$, $EF=3x + 10$, $DF=9x-39$, we set up the equation $47+(3x + 10)=9x-39$.
First, simplify the left - hand side: $47+3x + 10=57+3x$. So, $57+3x=9x-39$.
Subtract $3x$ from both sides: $57=9x-3x - 39$, which simplifies to $57 = 6x-39$.
Add 39 to both sides: $57 + 39=6x$, so $96 = 6x$.
Divide both sides by 6: $x = 16$. Then $EF=3x + 10=3\times16+10=48 + 10=58$.

Step4: Solve for $x$ in the fourth problem

Since $HI+IJ=HJ$, and $HI = 3x-5$, $IJ=x - 1$, $HJ=7x-27$, we set up the equation $(3x-5)+(x - 1)=7x-27$.
Simplify the left - hand side: $3x-5+x - 1=4x-6$. So, $4x-6=7x-27$.
Subtract $4x$ from both sides: $-6=7x-4x - 27$, which simplifies to $-6 = 3x-27$.
Add 27 to both sides: $-6 + 27=3x$, so $21 = 3x$.
Divide both sides by 3: $x = 7$.

Answer:

1. $LN = 37$
2. $MN = 23$
3. $EF = 58$
4. $x = 7$